# Problem

Consider the following grammar, where G is the initial symbol and {a,b,c,d,e} is the set of terminal symbols:

```O -> a
G -> F c | O c d | (eps)
F -> G b | O c F e
```
1. Examine the grammar and rewrite it so that an LL(1) predictive parser can be built for the corresponding language.
2. Compute the FIRST and FOLLOW sets for all non-terminal symbols in the new grammar and build the parse table.
3. Show the analysis table (stack, input, and actions) for the parsing process of the acbec input sequence.

# Solution

Something to keep in mind at all times: never eliminate the rules corresponding to the initial symbol

### Elimination of Mutual Recursion: Expanding F in G

Initial grammar:

```O -> a
G -> F c | O c d | (eps)
F -> G b | O c F e
```

Eliminating the singularity (O->a):

```G -> F c | a c d | (eps)
F -> G b | a c F e
```

Expanding F in G:

```G -> G b c | a c F e c | a c d | (eps)
F -> G b | a c F e
```

Eliminating left recursion in G:

```G -> a c F e c G' | a c d G' | G'
G' -> b c G' | (eps)
F -> G b | a c F e
```

Factoring G prefixes:

```G -> a c G" | b c G' | (eps)
G' -> b c G' | (eps)
G" -> F e c G' | d G'
F -> a c G" b | b c G' b | b | a c F e
```

Factoring F prefixes:

```G -> a c G" | b c G' | (eps)
G' -> b c G' | (eps)
G" -> F e c G' | d G'
F -> a c F' | b F"
F' -> G" b | F e
F" -> c G' b | (eps)
```

Eliminating non-terminal left corners (note that F becomes unreachable):

```G -> a c G" | b c G' | (eps)
G' -> b c G' | (eps)
G" -> a c F' e c G' | b F" e c G' | d G'
F' -> a c F' e c G' b | b F" e c G' b | d G' b | a c F' e | b F" e
F" -> c G' b | (eps)
```

Factoring F' prefixes, we get to the final version:

```G -> a c G" | b c G' | (eps)
G' -> b c G' | (eps)
G" -> a c F' e c G' | b F" e c G' | d G'
F' -> a c F' e F" | b F" e F" | d G' b
F" -> c G' b | (eps)
```

```FIRST(G)  = { a, b, (eps) }               FOLLOW(G)  = { \$ }